JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Vector Algebra

Consider the cube in the first octant with sides

O P, O Q

and

O R

of length 1, along the

x

-axis,

y

-axis and

z

-axis, respectively, where

O (0, 0, 0)

is the origin. Let

S (\frac{1}{2}, \frac{1}{2}, \frac{1}{2})

be the centre of the cube and

T

be the vertex of the cube opposite to the origin

O

such that

S

lies on the diagonal

O T

. If

p = S P, q = S Q, r = S R

and

t = S T

, then the value of

∣ (p \times q) \times (r \times t) ∣

is ____.

Answer: 0.5

Solution

The vertices are:

P (1, 0, 0)

Q (0, 1, 0)

R (0, 0, 1)

T (1, 1, 1)

, and

S (1/2, 1/2, 1/2)

p = \frac{i ^}{2} - \frac{j ^}{2} - \frac{k ^}{2}, q = - \frac{i ^}{2} + \frac{j ^}{2} - \frac{k ^}{2}

r = - \frac{i ^}{2} - \frac{j ^}{2} + \frac{k ^}{2}, t = \frac{i ^}{2} + \frac{j ^}{2} + \frac{k ^}{2}

Using the vector identity:

∣ (p \times q) \times (r \times t) ∣ = ∣ [p r t] q - [q r t] p ∣

Calculating the scalar triple products and simplifying:

= - \frac{4}{8} q - \frac{4}{8} p = \frac{4}{8} ∣ p + q ∣ = 0.5

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Numerical answer